Prove this inequation between two integrals

Question

I'm trying to prove this inequality but I'm completely stuck.

$$S_1=\int_0^1 s(u)du$$

$$S_2=\int_0^1 s(u)u \,du$$

Prove that:

$$S_1^2<2 S_2$$

$s(u)$ is monotonously decreasing and $s(0)=1$ and $s(1)=0$.

Answer 1

Example of monotonically decreasing functions: $-u^3, -arctan(u), e^{-u}$, etc.

Let's see if this inequality holds for certain examples of $s(u)$. Take $s(u)=1-u$

$S_1=\int_{0}^{1}s(u)du=1/2, S_2=\int_{0}^{1}s(u)udu=1/6$

$S_1^2=1/4, 2S_2=1/3 \implies S_1^2<2S_2$

Inequality did hold, so let's start the proof:

By integration by parts,

$S_2=\int_{0}^{1}s(u)udu = [uR(u)- R_2(u)]_{0}^{1} = R(1)-R_2(1)+R_2(0)=R(1)-R_2(1)$

$2S_2 = 2R(1)-2R_2(1)$

where $R(u)$ is the anti-derivative of $s(u)$. $R_2(u)$ is the anti-derivative of $R(u)$. We could remove $R_2(0)$ as we can define the anti-derivative function to have a constant such that $R(0),R_2(0)=0$. With this definition, we can decompose $S_1$ to be in terms of $R(u)$ as well by invoking the fundamental theorem of calculus for definite integrals

$S_1=\int_{0}^{1}s(u)=R(1)-R(0)=R(1)$

$S_1^2=(R(1))^2$

Now, we just have to compare $R^2_{1}(1)$ and $2R(1)-2R_2(1)$ which looks easier. It is time to make use of the 3 conditions provided. $s(0)=1,s(1)=0$ and $s(u)$ is monotonously decreasing. Something interesting we find out from these conditions is that $s(u)>0,0\le u < 1$. This means that $R(u),R_2(u)$ are monotonously increasing.

We can observe that since $R(u)<1$ for $u>0$. $y=s(u)$ and $y=1$ should only intersect once, at $u=0$. For $0<u\le 1,s(u)<1$ As both functions are all non-negative, and bounded in $0<y<1$, $\int_{0}^{1}s(u)du<\int_{0}^{1}1du \implies \int_{0}^{1}s(u)du<1$ and so $R(1)<1$

Since, $(R(1))^2<2R(1)$ as $0<R(1)<1$

$(R(1))^2<2R(1)+R_2(1)$

When we sub back $S_1(u), S_2(u)$

$S_1^2(u)<2S_2(u)$

Q.E.D

This was a fun exercise. I apologise if I made any mistakes along the way

Side note: this is not part of the problem but I found this interesting. Using a similar logic, since $R(u)$ is monotonously increasing, One can show $R(1)<R_2(1)$

For $0\le u<1$, $R(u)<R(1)$

$\int_{0}^{1}R(u)du<\int_{0}^{1}R(1)du \implies R(1)<R_2(1)$